The speed of the current in a river is 6 mph. A ferry operator who works that part of the river is looking to buy a new boat for his business. Every day, his route takes him 22.5 miles each way against the current and back to his dock, and he needs to make this trip in a total of 9 hours. He has a boat in mind, but he can only test it on a lake where there is no current. How fast must the boat go on the lake in order for it to serve the ferry operator’s needs?
In order to figure out how fast the boat must go on the lake, we first need to determine the speed of the boat relative to the water.
Let x be the speed of the boat in still water.
Against the current, the speed of the boat is x - 6 mph.
With the current, the speed of the boat is x + 6 mph.
Given that the distance is 22.5 miles each way and the total time is 9 hours, we can set up the equation:
22.5/(x-6) + 22.5/(x+6) = 9
Multiplying both sides by (x-6)(x+6) to eliminate the denominators, we get:
22.5(x+6) + 22.5(x-6) = 9(x-6)(x+6)
Solve for x:
22.5x + 135 + 22.5x - 135 = 9(x^2 - 36)
45x = 9x^2 - 324
9x^2 - 45x - 324 = 0
x^2 - 5x - 36 = 0
(x-9)(x+4) = 0
Since the speed of the boat cannot be negative, the only valid solution is x = 9.
Therefore, the speed of the boat on the lake must be 9 mph in order for it to serve the ferry operator's needs.